The pressure of water is 7.3851 kPa
<u>Explanation:</u>
Given data,
V = 150×

m = 1 Kg
= 2 MPa
= 40°C
The waters specific volume is calculated:
= V/m
Here, the waters specific volume at initial condition is
, the containers volume is V, waters mass is m.
= 150×
/1
= 0.15
/ Kg
The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15
/ Kg and 0.13
/ Kg.
= 350+(400-350) 
= 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
=
= 0.15
/ Kg
In saturated water labels for
= 40°C.
= 0.001008
/ Kg
= 19.515
/ Kg
The final state is two phase region
<
<
.
In saturated water labels for
= 40°C.
=
= 7.3851 kPa
= 7.3851 kPa
Answer:
#Initialise a tuple
team_names = ('Rockets','Raptors','Warriors','Celtics')
print(team_names[0])
print(team_names[1])
print(team_names[2])
print(team_names[3])
Explanation:
The Python code illustrates or printed out the tuple team names at the end of a season.
The code displayed is a function that will display these teams as an output from the program.
I’m crying looking at that.
Answer:
Yes
Explanation:
If the Ajax representative fails to correct the previous statement this can cause misrepresentation.
Answer:
74,4 litros
Explanation:
Dado que
W = nRT ln (Vf / Vi)
W = 3000J
R = 8,314 JK-1mol-1
T = 58 + 273 = 331 K
Vf = desconocido
Vi = 25 L
W / nRT = ln (Vf / Vi)
W / nRT = 2.303 log (Vf / Vi)
W / nRT * 1 / 2.303 = log (Vf / Vi)
Vf / Vi = Antilog (W / nRT * 1 / 2.303)
Vf = Antilog (W / nRT * 1 / 2.303) * Vi
Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25
Vf = 74,4 litros